'''
We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

If x == y, both stones are totally destroyed;
If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.
At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)



Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.


Note:

1 <= stones.length <= 30
1 <= stones[i] <= 1000
'''

class Solution(object):
    def lastStoneWeight(self, stones):
        """
        :type stones: List[int]
        :rtype: int
        """
        while len(stones) > 1:
            max_x = max(stones)
            stones.remove(max_x)
            max_y = max(stones)
            stones.remove(max_y)

            if max_x != max_y:
                stones.append(max_x-max_y)
        return stones[0] if stones else 0
